9x^2+18x-27=4x(x+3)

Solution for 9x^2+18x-27=4x(x+3) equation:

9x^2+18x-27=4x(x+3)

We move all terms to the left:

9x^2+18x-27-(4x(x+3))=0


We calculate terms in parentheses: -(4x(x+3)), so:

4x(x+3)

We multiply parentheses

4x^2+12x

Back to the equation:

-(4x^2+12x)


We get rid of parentheses

9x^2-4x^2+18x-12x-27=0

We add all the numbers together, and all the variables

5x^2+6x-27=0

a = 5; b = 6; c = -27;
Δ = b2-4ac
Δ = 62-4·5·(-27)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-24}{2*5}=\frac{-30}{10}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+24}{2*5}=\frac{18}{10}
=1+4/5
$